Answer:
V = 0.0806 m/s
Explanation:
given data
mass quarterback = 80 kg
mass football = 0.43 kg
velocity = 15 m/s
solution
we consider here momentum conservation is in horizontal direction.
so that here no initial momentum of the quarterback
so that final momentum of the system will be 0
so we can say
M(quarterback) × V = m(football) × v (football) ........................1
put here value we get
80 × V = 0.43 × 15
V = 0.0806 m/s
Barium cation has +2 charge and oxide anion has −2 charge
Answer:
0.28 m
Explanation:
The following data were obtained from the question:
Force (F) = 5×10¯⁶ N
Charge 1 (q₁) = 6.7×10¯⁹ C
Charge 2 (q₂) = 6.7×10¯⁹ C
Electrical constant (K) = 9×10⁹ Nm²C¯²
Distance apart (r) =?
Thus, the distance between the two charges can be obtained as follow:
F = Kq₁q₂/r²
5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²
5×10¯⁶ = 4.0401×10¯⁷ / r²
Cross multiply
5×10¯⁶ × r² = 4.0401×10¯⁷
Divide both side by 5×10¯⁶
r² = 4.0401×10¯⁷ / 5×10¯⁶
Take the square root of both side
r = √(4.0401×10¯⁷ / 5×10¯⁶)
r = 0.28 m
Therefore, the distance between the two charges is 0.28 m
-- "Free fall" means no forces acting on the object except for gravity.
-- An object dropped through air has two forces on it -- gravity and air resistance.
-- So we're looking for the part of the drop where air resistance is smallest.
-- The force of air resistance depends on the speed through the air, so the
force of air resistance is smallest when <em>velocity downward is smallest</em>.
Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
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If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
_____
<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.
